The solutions of the Schrödinger equation for a hydrogen atom have definite energy. Does this mean that they could be written as a superposition of plane waves of a single frequency - corresponding to that energy - with only the phases and directions differing?

2 Answers
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Let's consider a simpler system: 1D harmonic oscillator, i.e. system with potential $U=x^2$. The only plane waves with different directions and phases you can sum in 1D case is $e^{ikx}$ and $e^{-i(kx+\phi)}$. Now any way you try to sum these two plane waves, this will only give you a function, which oscillates at infinity. But there's no eigenstate here, which oscillates at infinity — all are bound. Thus they can't be represented as a linear combination of plane waves with single frequency.

In hydrogen there's also infinitely many bound states, which don't oscillate at infinity, and summing plane waves over sphere will give you spherical Bessel functions and the like, which do oscillate. So, the answer is no.

What you seem to confuse is frequency versus wave length. Plane waves are eigenstates of Hamiltonian only for a free particle. If a particle in state of plane wave is put in an inhomogeneous potential, it will no longer be in an eigenstate: its state will be a superposition of different energy states, thus there'll be no single frequency. What is a characteristic for a plane wave is its wavelength, which doesn't depend on potential (although the wave will be distorted in time evolution if it's not an eigenstate, and cease to be a plane wave).

Since hydrogen atom is characterized by Coulomb potential, which is non-constant, plane wave can't have single frequency in such a system — it'll distort in the next moment of time.

It's a bit naïve to generalise from 1D to 3D like that. Summing arbitrary directions while keeping only $|k|$ constant certainly does not always give periodic solutions, in fact integrating over the whole unit sphere should give something quite spherical-wave-like if I'm not mistaken.
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leftaroundaboutApr 29 '14 at 19:58

@leftaroundabout yeah, it'll give spherical Bessel functions if you choose correct phases. Anyway, OP's idea didn't work in 1D, so I just asserted that in general it won't work, including 3D.
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RuslanApr 29 '14 at 20:24

@Ruslan Could you elaborate on the harmonic oscillator? I think the plane waves you can use are actually $e^{i(kx+\phi'_j)}$ and $e^{i(kx+\phi_i)}$, with a free choice of phases $\phi'_j$ and $\phi_i$, not $e^{i(kx)}$ and $e^{-ikx+\phi}$.
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yippy_yayApr 29 '14 at 21:52

@SebastianHenckel well, one of these $\phi$ would be redundant, since it could be taken as a global phase factor, which doesn't change the state, what does change state is difference of these $\phi$s. The other one in my choice just should be imaginary. I'll fix.
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RuslanApr 30 '14 at 5:13

The solutions to the time-independent Schrödinger equation are independent of frequency, so your answer is trivially yes.

You might also ask whether the spatial parts of the wavefunction can be written as a superposition of plane waves $e^{ i\vec k \cdot \vec x}$ (they can, by Fourier decomposition) and whether the wavevectors $\vec k$ all have the same magnitude.

I had originally written that not all the plane wavevectors can have the same magnitude, since the bound wavefunctions are localized. A comment by Sebastian Henckel has made me realize this is not correct for plane wave which all have the same phase at the origin. However, I think it'd be quite challenging to produce the hydrogen radial wavefunctions using only plane waves which are coherent at the origin — in particular the feature that the $n$th radial wavefunction has exactly $n-1$ nodes at different values for $r$. A set of plane waves, in phase at the origin, will have infinitely many nodes in $r$, in spherical shells.

The physics here is that the states of the hydrogen which have definite energy do not have describe electrons with definite momentum, or definite wavenumber $|k|$.

I don't immediately see that the superposition of an infinite number of plane waves with wavevectors $\vec k$ of the same magnitude but different direction can't result in something which is localized.
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yippy_yayApr 29 '14 at 21:28